70 R 3-3 AND 71 I* ‘tcb4;oc;
70WASTEWATER CHARACTERISTICSr3-3 CHEMICALCHARACTERISTICS DEFINITIONAND APPLICATION71‘tcb4;oc;I *and the contaminants can be weighed. Pesticides and herbicides in concentrations o1 part per billion (ppb) and less can be determined accurately by several metho&including gas chromatography and electron capture or coulometric detectors [I 81.)f some of these compounds in the atmosphere may pose a significant public healthisk; and (3) they contribute to a general increase in reactive hydrocarbons in theitmosphere, which can lead to the formation of photochemical oxidants. The releasethese compourtds in sewers and at treatment plants, especially at the headworks,is of particular concern with respect to the health of collection system and treatmentplant workers. The release and control of VOCs is considered further in Chaps. 6 and9. The physical phenomena involved in the release of VOCs are considered in detailin Kef. 24.3fPesticides and Agricultural Chemicals. Trace organic compounds, such aspesticides, herbicides, and other agricultural chemicals, are toxic to most life forinsand therefore can be significant contaminants of surface waters. These chemicals arenot common coristituents of domestic wastewater but result primarily from surfacerunoff from agricultural, vacant, and park lands. Concentrations of these chemicalscan result in fish kills, in contamination of the flesh of fish that decreases their valueas a source of food, and in impairment of water supplies. Many of these chemicalsare also classified as priority pollutants.Measurement of Organic ContentOver the years, a number of different tests have been developed to determine theorganic content of wastewaters. In general, the tests may be divided into those usedto measure gross concentrations of organic matter greater than about 1 mg/L and thoseused to measure trace concentrations in the range of I O p L 2 to lop3 mg/L. Laboratorymethods commonly used today to measure gross amounts of organic matter (greaterthan 1 mg/L) in wastewater include: (1) biochemical oxygen demand (BOD); (2)chemical oxygen demand (COD); and (3) total organic carbon (TOC). Coinpleinentillgthese laboratory tests is the theoretical oxygen demand (ThOD), which is determinedfrom the chemical formula of the organic matter.Other methods used in the past included (1) total, albuminoid, organic, andammonia nitrogen, and (2) oxygen consumed. These determinations, with the exception of albuminoid nitrogen and oxygen consumed, are still included in completewastewater analyses. Their significance, however, has changed. Whereas formerlythey were used almost exclusively to indicate organic matter, they are now used todetermine the availability of nitrogen to sustain biological activity in industrial wastetreatment processes and to foster undesirable algal growths in receiving water.Trace organics in the range of lo-’* to IOp3 mg/L are determined using instrumental methods including gas chromotography and mass spectroscopy. Within thepast 10 years, the sensitivity of the methods used for the detection of trace organiccompounds has improved significantly and detection of concentrations in the range of1 0 - mg/L is now almost a routine matter. The concentration of pesticides is typically measured by the carbon-chloroformextract method, which consists of separating the contaminants from the water bypassing a water sample through an activated-carbon column and then extracting thecontaminant from the carbon using chloroform. The chloroform can then be evaporated2Biochemical Oxygen Demand. he most widely used parameter of organic poltion applied to both wastewater and surface water is the 5-day BOD (BOD5). Thirdetermination involves theganisms in the biochemicalof the BOD test, it has a number of limitations (which are discussed later in thicsection). It is hoped that, through the continued efforts of workers in the field, oncof the other measures of organic content, or perhaps a new measure, will ultimatel!be used in its place. Why, then, if the test suffers from serious limitations, is furthespace devoted to it in this text? The reason is that BOD test results are now use((1) to determine the approximate quantity of oxygen that will be required to biologicallyLtabilize the organic matter present, (2) to determme me size ot waste treatmen-ties,and (3) to measure the efficiency of some treatment processes, and (4) tcdetermine compliance with wastewater discharge permits. Because it is likely that thtBOD test will continue to be used for some time, it is important to know as much apossible about the test and its limitations.To ensure that meaningful results are obtained, the sample must be- suitah11-.so that adequate nutrients and oxygeiormally, several dilutions are prepare(alues. The ranges of BOD that can bcmeasured with various dilutions based on percentage mixturesand direct pipettin)are reported in Table 3-10. The general procedure for preparing the BOD bottles foincubation is illustrated in Fig. 3-11.When the samplelarge population of microorganisms (untreatecwastewater, for example)not necessary. If required, the dilution water i“seeded” with a bacterialbeen acclimated to the organic matter or othematerials that may be present in the wastewater. The seed culture that is used to preparcthe dilution water for the BOD test is a mixed culture. Such cultures contain largcnumbers of saprophytic bacteria and other organisms that oxidize the organic matterIn addition, they contain certain autotrophic bacteria that oxidize noncarbonaceoumatter. A variety of commercial seed preparations are also available.The incubation period is usually five days at 20 C, but other lengths of timcand t e m p e G e s can be used. Longer time periods (typically seven days), whiclcorrespond to work schedules, are often used, especially in small plants where thclaboratory staff is not available on the weekends. The temperature, however, shoulcbe constant throughout the test. The dissolved oxygen of the samples is measured (secFig. 3-12) before and after incubation, and the BOD is calculated using Eq. 3-2 oEq. 3-3.When dilution water iBOD, mg/L D I - 0P2(3-2
.WASTEWATER CHARACTERISTICS3-3 CHEMICAL CHARACTERISTICS: DEFINITION AND APPLICATION3LE 3-10Essentialnutrients(N, P, K, Fe, etc.)and otheradditivesID measurable with various dilutionssamDlesa\By direct pipetting intoBODmixtureRange 0,000-70,00010,000-35,0004,000-14,0002,000 -7,0001,000-3,500400-1,400200-700100-35040-1 4020-7010-354-1 .0300.0Range ofI/Test (waste) sample, b' ,containing organicmatter and anadequate number ofbacteria (volume of testwater300 mL bottlesy using percent mixturesAir73sample depends onestimated 30-1 0512-426-210-7Glass stoppered BODbottle (volume 300 mL)Unseededdilutionwaterwith test sample plusunseeded dilution water(unseeded test sample)(a)EssentialAirTest (waste) sample, 4 ,containing organicmatter and nobacteria or a limitednumber of bacteriaRef. 32.-EIIWhen dilution water is seeded,(3-3)Seededdilution"here D dissolved oxygen of diluted sample immediately after preparation, mg/LD2 dissolved oxygen of diluted sample after 5 d incubation at 2O"c, mg/LP decimal volumetric fraction of sample usedB 1 dissolved oxygen of seed control before incubation, mg/LB2 dissolved oxygen of seed control after incubation, mg/Lf ratio of seed in sample to seed in control (% seed in D1)/(%seed in B1)Biochemical oxidation is a slow process and theoretically takes an infinite timeto go to completion. Within a 20-day period, the oxidation of the carbonaceous organicmatter is about 95 to 99 percent complete, and in the 5-day period used for the BODtest, oxidation is from 60 to 70 percent complete. The 20 C temperature used is anaverage value for slow-moving streams in temperate climates and is easily duplicatedin an incubator. Different results would be obtained at different temperatures becausebiochemical reaction rates are temperature-dependent .The kinetics of the BOD reaction are, for practical purposes, formulated inaccordance with first-order reaction kinetics and may be expressed asL E &A(3-4)L W L ? LTtykk.te twaterBOD bottle filledwith seededdilution water(seeded blank)BOD bottle filledwith test sample plusseeded dilution water(seeded test sample)FIGURE 3-11Procedure for setting up BOD test bottles: (a)with unseeded dilution wafer and (b) with seeded dilution water [23].where L t is the amount of the first-stage BOD remaining in the water at timek is the reaction rate constant. This equation can be integrated asfand(3-6)where &or BODL is the BOD remaining at time t 0 (i.e., the total or ultimatefirst-stage BOD initiany present). The relation between k (base e) and K (base 10) isas follows:K(base 10) k( base e )2.303(3-7)
3-3 CHEMICAL CHARACTERISTICS:DEFINITION AND APPLICATIONWASTEWATER CHARACTERISTICS75Example 3-2 Calculation of BOD. Determine the 1-day BOD and ultimate first-stage BODfor a wastewater whose 5-day, 20 C BOD is 200 mg/L. The reaction constant k(base e) 0.23d-' .Solution1. Determine ultimate BOD. -Z F KL, 200 - 5-c&e-kry s L,-eL L ( 1 - e-kr)Lo(' - e - 5 ( 0 2 3 ) ) LJI -0.316)LD133*)D#SI25%XI3Lo 293 mg/L2. Delerminc I-day BOD.L, L1 293(e-023'") 293(0.795) 233 mg/LYI 6-L I 293 - 233 60 mg/L2 0 C ) m d-'varies significantly,.05 to 0.3 d-' (baseary with time andt'IGURE 3-12neasurement of oxygen in BOD bottle with aDO probe equipped with a stirring mechanism.wastewater sample isble, however, to determine the reaction constantThe following approximate equation, which isThe amount of BOD remaining at time t equals(3-8)t any time t , equalsmd y , the amount of BOD that has beey r L,- L , LO '- e - k t )aNote that the 5-diiy BOD equals95 L -L G I - e-5k)(3-9)-k r k20e(T-20)--(3-10)7-EDc--This relationship is shown in Fig. 3-13. The use of the BOD equations is illustratedin Example 3-2.(3-1 1)& .W#X-Xfrom 0 to tt(l3e,23.or-e33ca t timeTime. trgo- 3L-8./tar&4-34/Pi35--oBOD remainingTime, daysFIGURE 3-13Formation of the first-stage BOD curve.FIGURE 3-14Effect of the rate constant k on BOD (for a given L value).
i.3-3 CHEMICAL CHARACTERISTICS: DEFINITION AND APPLICATIONWASTEWATER CHARACTERISTICSLe value of 8 has been found to vary from 1.056 in the temperature range betweenand 30 C to 1.135 in the temperature range between 4 and 20 C [15]. A value ofoften quoted in the literature is 1.047 [ 141, but it has been observed that this valueies not apply at cold temperatures (e.g., below 20 C) [15].Nitrification in the BOD test. Noncarbonaceous matter, such as ammonia,produced during the hydrolysis of proteins. Two groups of autotrophic bacteria arelpable of oxidizing ammonia to nitrite and subsequently to nitrate. The generalizedactions are as follows:HNO2 H20(3-12)nitrite-forming bacteriaNH3 1 0 2(3-13)nitrate-forming bacteriaHN03,)H N O io2NH3 t-202HN03 H20 (3-14)to nitrate is calledhe oxygen demand associated with theexertion of thele nitrogenous biochemical oxveenx-2va--m n demandin a BOD test for- --,ecause the reproductive rate of the nitrifying bacteria is slow, it normP1L J takes fromto 10 days for them to reach significant numbers and to exert a mea -1. able oxygenemand. However, if a sufficient number of nitrifying bacteria are pi' :nt initially,le interference caused by nitrification can be significant.When nitrification occurs in the BOD test, erroneous interpretations of treatnent operating data are possible. For example, assume that the effluent BOD frombiological treatinent process is 20 mg/L without nitrification and 40 mg/L withiitrification. If the influent BOD to the treatment process is 200 mg/L, then the cor-Where a sufficient number of nitrifyingorganisms are present, nitrificationcanoccur as shown by the dotted curve.2oxygen demand, NBODdmDCt/Nitrificationis usually observed tooccur from 5 to 8 days after thestart of the BOD incubationperiod.vCarbonaceous biochemicaloxygen demand, BOD orCBODTime, dFIGURE 3-15Definition sketch for the exertion of the carbonaceous and nitrogenous biochemical oxygen demandin a waste sample.77responding BOD removal efficiency would be reported as 90 and 80 percent withoutand with nitrification, respectively. Thus, if nitrification is occurring but is notsuspected, it might be concluded that the treatment process is not performing wellwhen in actuality it is performing quite well.Carbonaceous biochemical oxygen demand (CBOD). The interferencecaused by the presence of nitrifying bacteria can be eliminated by pretreatmen of the sample or%y the use of inhibitory agents. Pretreatment procedures includepasteurization, chlorination, and acid treatment. Inhibitory agents are usually chemicalin nature and include compounds such as methylene blue, thiourea and allylthiourea,2-chlor-6 (trichloromethyl) pyridine, and other proprietary products [37]. Suppressionof the nitrification reaction in the BOD test is listed as a standard procedure in thelatest edition of Standurd Methods [18]. The results of the suppressedBOD test shouldbe reportedas CBOD (carbonaceous biochemical oxygen demand). The CBOD test is. .now being used as substitue for the BOD test in discharge permits, especially wherenitrification is known to occur.-Analysis of BOD data. The value of k is needed if the BODS is to be usedto obtain L , the ultimate or 20-day BOD. The usual procedure followed when thesevalues are unknown is to determine k and L from a series of BOD measurements.There are several ways to do this, including the method of least-squares, the methodof moments [ 111, the daily-difference method [27], the rapid-ratio method [ 161, theThomas method [26], and the Fujimoto method [5]. The least-squares method andthe Fujimoto method are illustrated in the following discussion.The least-squares method involves fitting a curve through a set of data points,so that the sum of the squares of the residuals (the difference between the observedvalue and the value of the fitted curve) must be a minimum. Using this method, avariety of different types of curves can be fitted through a set of data points. Forexample, for a time series of BOD measurements on the same sample, the followingequation may be written for each of the various n data points: k(L-y,)(3-15)dt t nIn this equation both k and L are unknown. If it is assumed that dyldt represents thevalue of the slope of the curve to be fitted through all the data points for a given kand L value, then because of experimental error, the two sides of Eq. 3-15 will notbe equal but will differ by an amount R. Rewriting Eq. 3-15 in terms of R for thegeneral case yieldsdYR k(L-y) -dtSimplifying and using the notation y' for dyldt givesR k L - ky - y 'Substituting a for kL and -b for k givesR a -k by - y '(3-16)(3-17)(3-18)
‘83-3 CHEMICAL CHARACTERISTICS: DEFINITION AND APPLICATIONWASTEWATER CHARACTERISTICS, -?ow, if the sum of the squares of the residuals R is to be a minimum, the followingquations must hold:(3- 19)d-xdbR2 1 2R-ddbR 0The slope y’ is computed as follows:dY y ’ 4a75af the indicated operations in Eqs. 3-19 and 3-20 are carried out using the value of(3-21) bC(3-22)UC yy2 - 2 yy’ 0Ya 7.5 and b -0.271k -0 0.271 (base e )U7.5 27.7 mg/L-0.2714. Prepare an arithmetic plot of BOD, I versus BOD,, and on the same plot draw a line with itslope of 1 . The value at the intersection of the two lines (BOD 27.8 mglL) corre\pond\to the ultimate BOD.L -- -5.0---rExample 3-3 Calculation of BOD constants using the least squares and theFujimoto methods. Compute L and k using the least-squares and Fujimoto methods for thefollowing BOD data reported for a stream receiving some treated effluent:1 2 14 16 18yYZ24’6811182224-4.501212.753241.50484576 1.00751,505Y’9.754.0?3.0,-- 8m2.0-1 .o001020304050BOD f , mg/L10Solution1. Set up a computation table and perform the indicated steps.Time 1505b - 156.0 03. Determine the values of k and L.- Yn l -Yn-IApplication of the least-squares method in the analysis of BOD data is illustrated inExample 3-3, which follows the discussion of the Fujimoto method.In the Fujimoto method [ 5 ] , an arithmetic plot is prepared of BOD, l versusBOD,. The value at the intersection of the plot with a line of slope 1 correspondsto the the ultimate BOD. After the BODL has been determined, the rate constant isdetermined using Eq. 3-9 and one of the BOD values. The application of the Fujimotomethod is illustrated in Example 3-3.t, d 750 - 9.75 0bwhere n number of data pointsu -bLb -k(base e)L -dby Yr, mglLIYn l -Yn-l2at2. Substitute the values computed in step 1 in Eqs. 3-21 and 3-22, and solve for u and 6 .(3-20)he residual R defined by Eq. 3-18, the following set of equations result:n a b y - y ’ O79YY’49.549.533.024.0156.0-k 0.293 d-’Respirometric determination of BOD. Determination of the BOD valueand the corresponding rate constant k can be accomplished more efficiently in thelaboratory using an instrumented large-volume (1 .O L) electrolysis cell or a laboratoryrespirometer. An electrolysis cell (see Fig. 3-16u) may also be used to obtain a continuous BOD [38,39].Within the cell, oxygen pressure over the sample is maintainedconstant by continually replacing the oxygen used by the micoorganisms. Oxygen
WASTEWATER CHARACTERISTICS3-3 CHEMICAL CHARACTERISTICS: DEFINITION AND APPLICATIONOxvaen electrodeUltimate carbonaceous biochemical oxygendemand of waste sample (BOD,)/Total amount of organic matterremaining in BOD bottle//Iform of cell tissue:ionLogLagphase4growthPhase- --Time, f---BOD,(bi\,BODr (remaining)- Stationary growth phaseDeath phase(a181Amount of organic matter remaining(actual and idealized)wIdealized BOD curveIGURE 3-16and (b) commercial respiromlectrolytic respirometerfor BOD determination: (a) schematic [37,38]ter with multiple electrolysis cells.eplacement is accomplished by means of an electrolysis reaction in which oxygen isxoduced in response to changes in the pressure. The BOD readings are determined)y noting the length of time that the oxygen was generated and by correlating it tohe amount of oxygen produced by the electrolysis reaction. Advantages of the:lectrolysis cell over a conventional laboratory respirometer are that (1) the use of alarge (1 L) sample minimizes the errors of grab sampling and pipetting in dilutions, and:2) the value of the BOD is available directly. A typical example of a commerciallyavailable electrolytic respirometer with multiple electrolysis cells is also shown inFig. 3-16b.Time, f(4Limitations in the BOD test. The limitations of the BOD test are asfollows: (1) a high concentration of active, acclimated seed bacteria is required; (2)pretrqatment is needed when dealing with toxic wastes, and the effects of nitrifyingorganisms must be reduced; ( 3 ) only the biodegradable organics are measured; (4)the test does not have stoichiometric validity after the soluble organic matter presentin solution has been used (see Fig. 3-17); and (5) an arbitrary, long period of timeFIGURE 3-17Functional analysis of the BOD test: (a) interrelationship of organic waste, bacterial mass (cell tissue), total organic waste, and oxygen consumed in BOD test and (b) idealized representation of theBOD test [23].
2WASTEWATER CHARACTERISTICS3-3 CHEMICALCHARACTERISTICS:DEFINITIONAND APPLICATION.,required to obtain results. Of the above, perhaps the most serious limitation is thate 5-day period may or may not correspond to the point where the soluble organicatter that is present has been used. The lack of stoichiometric validity at all timesduces the usefulness of the test results.hemical Oxygen Demand. The COD test is used to measure the content of-ganic matter of both wastewater and natural waters. The oxygen equivalent ofie organic matter that can be oxidized is measured by using a strong chemicalitidizing agent in an acidic medium. Potassium dichromate has been found to beitcellent for this purpose. The test must be performed at an elevated temperature. Aatalyst (silver sulfate) is required to aid the oxidation of certain classes of organicDmpounds. Since some inorganic compounds interfere with the test, care must beiken to eliminate them. The principal reaction using dichromate as the oxidizinggent may be represented in a general way by the following unbalanced equation:Organic matter (c,H o,) Cr20q2 H Cr 3 COz H20(3-23)The COD test is also used to measure the organic matter in industrial andnunicipal wastes that contain compounds that are toxic to biological life. The CODIf a waste is, in general, higher than the BOD because more compounds can behemically oxidized than can be biologically oxidized. For many types of wastes, its possible to correlate COD with BOD. This can be very useful because the COD:an be determined in three hours, compared with five days for the BOD. Once the:orrelation has been established, COD measurements can be used to good advantage'or treatment-plant control and operation.83Example 3-4 Calculation Of ThOD. Determine the ThOD for glycine (CH2(NH2)COOH)using the following assumptions:1. In the first step, the organic carbon and nitrogen are converted to carbon dioxide (COZ)andammonia (NH,), respectively.2 . In the second and third steps, the ammonia is oxidized sequentially to nitrite and nitrate.3. The ThOD is the sum of the oxygen required for all three steps.Solution1. Write balanced reaction for the carbonaceous oxygen demand.CH2(NH2)COOH :02- NH3 2C02 H202. Write balanced reactions for the nitrogenous oxygen demand.(a) 1202NH3 HN02 H20 202NH3 HN03 H203. Determine the ThOD.ThOD (3 !)mol 02/mol glycine 3.5 mol 02/mol glycine X 3 2 g/mol 0, 112 g 02/mol glycinerota1 Organic Carbon. Another means for measuring the organic matter presentin water is the TOC test, which is especially applicable to small concentrations ofxganic matter. The test is performed by injecting a known quantity of sample intoB high-temperature furnace or chemically -oxidizing environment. The organic carbonis oxidized to carbon dioxide in the presence of a catalyst. The carbon dioxide that isproduced is quantitatively measured by means of an infrared analyzer. Acidificationand aeration of the sample prior to analysis eliminates errors due to the presence ofinorganic carbon. If VOCs are known to be present, the aeration step is omitted toeliminate their removal by stripping. The test can be performed very rapidly and isbecoming more popular. Certain resistant organic compounds may not be oxidized,however, and the measured TOC value will be slightly less than the actual amountpresent in the sample. Typical TOC values for wastewater are reported in Table 3-16in Sec. 3-5.Theoretical Oxygen Demand. Organic matter of animal or vegetable origin inwastewater is generally a combination of carbon, hydrogen, oxygen, and nitrogen.The principal groups of these elements present in wastewater are, as previously noted,carbc,hydrates, proteins, fats, and products of their decomposition. The biologicaldecomposition of the substances is discussed in Chap. 8. If the chemical formula ofthe organic matter is known, the ThOD may be computed, as illustrated in Example3-4.Correlation Among Gross Measures of Organic Content. Establishment ofconstant relationships among the various measures of organic content depends primarily on the nature of the wastewater and its source. Of all the measures, the mostdifficult to correlate to the others is the BOD5 test, because of the problems citedpreviously (see BOD discussion). For typical untreated domestic wastes, however,the BODs/COD ratio varies from 0.4 to 0.8, and the BODs/TOC ratio varies from1.O to 1.6. It should also be noted that these ratios vary considerably with the degreeof treatment the wastewater has undergone. Because of the rapidity with which theCOD, TOC, and related tests can be conducted, it is anticipated that more use willbe made of these tests in the future.Inorganic MatterSeveral inorganic components of wastewaters and natural waters are important inestablishing and controlling water quality. The concentrations of inorganic substancesin water are increased both by the geologic formation with which the water comes incontact and by the wastewaters, treated or untreated, that are discharged to it [17,20].The natural waters dissolve some of the rocks and minerals with which they comein contact. Wastewaters, with the exception of some industrial wastes, are seldom
WASTEWATER CHARACTERISTICS 3-3 CHEMICAL CHARACTERISTICS: DEFINITION AND APPLICATION 75 'IGURE 3-12 neasurement of oxygen in BOD bottle with a DO probe equipped with a stirring mechanism. The amount of BOD remaining at time t equals (3-8) (3-9) (3-10) md y, the amount of BOD that a has bee t any time t, equals yr L,- L, LO ' - e-kt) Note that the 5-diiy BOD equals 95
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